Question

Find the equation of hyperbola whose foci are (±5,0) and the transverse axis is of length 8

Answer 1

Foci are S(±5,0) ∴ ae = 5

Length of transverse axis = 2a = 8, a = 4 e = 5/4 b2 = a2 (e2 – 1) = 16(25/6 - 1) = 9

Equation of the hyperbola is x2/16 – y2/9 = 1 9x2 – 16y2 = 144

Answer 2

≡QUESTION≡

Find the equation of hyperbola whose foci are (±5,0) and the transverse axis is of length 8.

                                                           

║⊕ANSWER⊕║

Foci (±5, 0), the transverse axis is of length 8.

Foci is on the x-axis.

∴The equation of the hyperbola is of the form

$\frac{x^{2} }{a} - \frac{y^{2} }{b} = 1$

Since the foci are (±5, 0), c = 5.

Since the length of the transverse axis is 8, 2a = 8 ⇒ a = 4.

We know that a² + b² = c²

∴4² + b² = 5²

⇒ b² = 25 – 16

        = 9

Thus, the equation of the hyperbola is

$\frac{x^{2} }{16} - \frac{y^{2} }{9} = 1$